The Help of Derivatives - Assignment Example

?Pre-Calculus Module 5 Number Basic applicable rules If f(x) = ex, then f '(x) = ex 2) If f(x) = eg(x), then f '(x) = g'(x).eg(x) 3) If f(x) = ln x, then f '(x) = 1/x   (x > 0) 4) If f(x) = ln g(x), then f '(x) = g'(x)/g(x)    [g(x) > 0] (Tan, 2012). 1. Find the derivatives for the following functions. a. f(X) =100e10x f ' (x) = 1000 e10x b. f(X) = e (10X-5) Let u = 10x – 5 and y = eu Using chain rule to obtain the derivative of the function, f '(x) = (dy / du) (du / dx) dy / du = eu du / dx = 10 f '(x) = (e u)(10) = 10 e u Substituting u = 10x – 5 into the equation, f '(x) = 10 e (10x – 5) c. f(X) =ex3 f '(x) = 3x2 ex3 d. f(X) =2X2 e (1-X2) r (x) = 2x2, r' (x) = 4x s(x) = e (1-X2) s' (x) = -2 e (1-X2) Applying product rule, f '(x) = 2X2-2 e (1-X2) + 4x e (1-X2) = -8x2 e (1-X2) + 4x e (1-X2) = (4x -8x2) e (1-X2) e. f(X) =5X e (12-2x) Let r(x) = 5x, r' (x) = 5 and s(x) = e (12-2x), s' (x) = -2 e (12-2x) f '(x) = 5x (-2 e (12-2x)) + 5 e (12-2x) f '(x) = -10x e (12-2x) + 5 e (12-2x) = (-10 + 5) e (12-2x) f. f(X) =100e(x8 + x4) f '(x) = 8x7 + 4x3e(x8 + x4) g. f(X) = e (200X-X2 + x^100) f '(x) = 200 – 2x + 100x^99 e (200X-X2 + x^100) 2. Find the derivatives for the following functions: a. f(X) = ln250X b. f(X) = ln (20X-20) c. f(X) = ln (1- X2) d. f(X) = ln (5X + X-1) e. f(X) = Xln (12- 2X) f. f(X) = 2Xln(X3 + X4) g. f(X) = ln (200X - X2 + X100) Solutions The derivative of the function y = ln x is obtained by d/dx (ln x) = 1/x. d/dx logex = 1/x, suppose y = ln x, then dy/dx = 1/x a. f(X) = ln250X log ab = log a + log b Therefore, the equation can be rewritten as f (x) = ln 250 + ln x d/dx ln 250 = 0 (derivative of a constant) d/dx (ln x) = 1/x Hence dy/dx = 1/x. b. f(X) = ln(20X-20) If y = ln u and u is some function of x, then dy/dx = u'/u If y = ln f(x), then dy/dx = f ' (x)/ f(x) Let u = 20x – 20 u' = 20 dy/dx = 1. u'/u = 20/(20x – 20) c. f(X) = ln (1- X2) Let u = (1- X2) Then u' = -2x dy/dx = 1. u'/u = -2x/(1- X2) d. ln(5X + X-1) Let u = (5X + X-1) Then u' = 5 – x-2 Therefore, dy/dx = 1. u'/u = 5 – x-2/ (5X + X-1) e. f(X) = X ln(12- 2X) Using rule 4 above and product rule, r(x) = x r' (x) = 1 s(x) = ln (12- 2X) s' (x) = -2/ (12 – 2x) Therefore, f ' (x) = x (-2/ (12- 2x) + 1. ln (12-2x) f ' (x) = 2x/ (12 – 2x) + ln (12-2x). f. f(X) = 2X ln(X3 + X4) Let r(x) = 2x, therefore, r' (x) = 2 Similarly, if s(x) = ln (X3 + X4), s'(x) = (3x2 + 4x3)/ (X3 + X4) Therefore, f ' (x) = 2x ((3x2 + 4x3)/ (X3 + X4)) + 2 ln (X3 + X4) g. f(X) = ln(200X - X2 + X100) u = ln (200X - X2 + X100) u' = 200 -2x + 100x99 f ' (x) = dy/dx = u'/u = 200 -2x + 100x99/ (200X - X2 + X100) 3. Find the indefinite integrals for the following functions a. f(X) = e6X = ? e6X dx = 1/6e6X + C b. f(X) = e (5X-5) = 5/2 x2-5x e (5X-5) c. f(X) = 5eX = ? 5eX dx = 5 ? eX dx = 5eX + C d. f(X) = 1/ (1 + X) = ln ?1 + x? + C e. f(X) = 5/X = 5 integral [1/x] dx = 5 ln ?x?+ C 4. Find the definite integrals for the following functions a. f(X) = e2X over the interval [2, 4] =Integral 42 [ e2x ] dx = [ 1/2 e2 ( 4) + C ] - [ 1/2 e2 ( 2 ) + C ] = 1/2 [ e8 - e4 ] b. f(X) = 2eX over the interval [0, 2] =Integral 20 [2eX] dx = [e2 + C] - [e0 + C] = e2 – e0 d. f(X) = 2/ (2 + X) over the interval [2, 5] Let u = 2 + x, when x = 2, u = 2 + 2 = 4 and when x = 5, u = 2 + 5 = 7 = ln [?2 + x?] 52 = ln (7) – ln (2) e. f(X) = 10/X over the interval [3, 10] =dx = 10 integral [ 10 / x ] dx = 10 [ ln | x | ] + C, so Integral103 [ 10/ x ] dx = [ 10 ln | 10 | + C ] - [ 10 ln | 3 | + C ] = 10 ln 10 – 10 ln3 = 10 [ln10 – ln 3] Part2: Application of Calculus in Business Decision-Making Calculus is extensively used in making business decisions, which are critical for the success and survival of every business enterprise. Derivatives have wide applications in the business world. Derivatives are used to measure rate of change of a function in relation to the changes in variables (inputs) under focus. At some given value of an input, the derivative tells us the linear estimate of the function, which is close to the value. This idea is applicable in determination of optimization levels in businesses, stock curves and other critical success factors. Business people usually pursue to optimize and maximize on various business opportunities by maximizing profits and minimizing costs. Without the help of derivatives, this would be practically impossible. As such, calculus is an integral part of business and is an indispensable element in the world of business. Calculus enables business people to determine the price that maximizes revenue. It makes it possible for business community to determine effectively the effect of increasing profit maximizing price on sales and total revenue of the firm. Similarly, calculus permits business people to predetermine the effect of reducing prices of commodities on sales and total revenue of the firm. It allows operation managers to forecast operation costs effectively as well as forecast the demand for their company’s products. The maximum and minimum points in a curve (when gradient = 0), help in estimating inputs when the function is at its optimal level. Integration techniques allow companies to determine marginal costs and revenues resulting from change in operations of the firm. Marginal analysis assist in determining break-even-points and profit maximizing prices and output. Profit maximizing output occurs where marginal revenue equals to marginal cost (Hall & Lieberman, 2010; Hoag & Hoag, 2006). Derivatives can be used to determine additional revenue (marginal revenue) to be expected as a result of using additional resources (input). Total revenue is obtained by multiplying price charged of an item by the quantity available for sale. Therefore, applying product rule, we can obtain marginal revenue by multiplying the derivative of price of the product by price demanded plus quantity available for sale. Total revenue curve and total cost curve of a company depicts changes in total revenue and total cost respectively in relation to variation in level of production. Marginal analysis is used to determine the marginal cost of labor as well as its marginal revenue. Profit maximizing rule as regards labor (variable input) is given by MRPL = MCL, where L is labor. The marginal MRP is the change in TR per unit change in input. MRPL = ?TR/?L. MRPL = MR x MPL (Carbaugh, 2011). Calculus is crucial for making production decisions. For instance, when manufacturing metal discs, it is usually important to determine the error of tolerance of the material used. This is made possible with the idea of calculus (function and limit). As such, it would be difficult for manufacturers to make accurate critical decision during production process without applying the calculus mathematics. Calculus is also used in making borrowing decisions and hedging by firms. Companies are usually faced with the challenge of fluctuating interest rates on borrowing and stand to loss if inappropriate decision is made. Financial managers apply the technique of calculus to ensure that the firm only goes for what it needs most. Calculus helps companies to spend reasonably on their acquisitions (Hoag & Hoag, 2006). Graph Of Profit Maximization Using Marginal Approach The profit is maximized when: , which is equivalent to the point where MR = MC, which is at point A in the graph. References Carbaugh, R.J. (2011). Contemporary economics: an applications approach. Armonk, N.Y.: M.E. Sharpe. Hall, R.E. & Lieberman, M. (2010). Microeconomics: principles and applications. Mason, OH: South-Western, Cengage Learning. Hoag, A.J. & Hoag, J.H. (2006). Introductory economics. London: World Scientific, cop. Tan, S.T. (2012). Applied calculus for the managerial, life, and social sciences: a brief approach. Belmont, CA: Brooks/Cole, Cengage Learning. Read More